Question

If you dropped a rock and it fell 4.9 m in 1 s , how far would it fall in 3 s

Answer 1

The distance "d" that it has fallen is proportional to the square of the time "t" since you dropped it:
`d = k * t^(2)` for some constant "k".

The questions asks: If you dropped a rock and it fell 4.9m in 1s, how far would it fall in 3s?

Changing "x" by a factor of 'c' changes "y" by a factor of c^2.

`x^(c) = c^(2) = y`

For example:
`x^(2) = 2^(2) = y`

So:

d = 4.9 * (3)^2

d = 44.1m

Your final answer is: 44.1m

Answer 2

The formula relevant for this is:

h = v0t + 0.5 gt^2

since the rock was dropped, therefore:

h = 0.5 gt^2

we can see that:

h / t^2 = 0.5 g = constant

therefore:

4.9 m / (1 s)^2 = h / (3 s)^2

h = 44.1 m